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Q.
Two spheres of electric charges $+2\, nC$ and $-8\, nC$ are placed at a distance $'d'$-apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?
In the first condition,
Given, $q_{1}=+2 n C=2 \times 10^{-9} C$
$q_{2} =-8 \,nC $
$=-8 \times 10^{-9} \,C $
$F =\frac{k q_{1} q_{2}}{r^{2}} $
$F =\frac{k\left(2 \times 10^{-9}\right) \times\left(8 \times 10^{-9}\right)}{d^{2}} $
$F =\frac{k\left(16 \times 10^{-18}\right)}{d^{2}}\,\,\,\,\dots(i)$
In the second condition,
$F_{1}=\frac{k q_{1} q_{2}}{d' 2}$
After touching of sphere to each other the total charge $=2-8=-6 \,nC$
Charge on each sphere $=3\, nC$
$=3 \times 10^{-9} \,C$
$F_{1}=\frac{k\left(3 \times 10^{-9}\right)\left(3 \times 10^{-9}\right)}{\left(d'\right)^{2}}$
$F_{1}=\frac{k\left(9 \times 10^{-18}\right)}{\left(d'\right)^{2}}$
From Eqs. (i) and (ii), we get
$\frac{k\left(16 \times 10^{-18}\right)}{d^{2}}=\frac{k\left(9 \times 10^{-18}\right)}{\left(d'\right)^{2}}$
$ \frac{d^{2}}{d^{2}} =\frac{16 \times 10^{-18}}{9 \times 10^{-18}} $
$\frac{d^{2}}{d' 2} =\frac{16}{9}$
$ \frac{d}{d'} =\sqrt{\frac{16}{9}} $
$ \frac{d}{d'} =\frac{4}{3} d '$
$=\frac{4}{3} d'$
$ d' =\frac{3}{4} d $